sub wanted_c
                {
                   if ((-d $_) && ($d_mode)
                   {
                        change_mode($_, $d_mode);
                        $d_mode = "";
                        wanted_c();
                    }
                    elsif ((-d $_) && ($d_own))
                    {
                        change_owner($_, $d_own);
                        $d_own = "";
                        wanted_c();
                    }
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That way it gets called as many times as is applicable to the file. Thanks for the pointer - I look forward to looking into your hash idea, which is slightly coalescing in my brain, but wouldn't mind your nudging it a bit with some advice/examples.

-Shane UPDATE: This fixed the problem but created another - it is doing everything, but only on one file! I see it is because of setting the variables to "", but am kinda unsure of what to do instead...

2nd UPDATE: Things are good now except for one thing, in order to fix the problem I created

my %info = %{ $config{$test}{$tester} };
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after the spot you suggest a hash. I then put

                else
                {
                    while(my($key,$val) = each(%info))
                    {
                        if ($key eq "dir-mode") { $d_mode = $val}
                        elsif ($key eq "file-mode") { $f_mode = $val}
                        elsif ($key eq "link-mode") { $l_mode = $val}
                        elsif ($key eq "dir-owner") { $d_own = $val}
                        elsif ($key eq "file-owner") { $f_own = $val}
                        elsif ($key eq "link-owner") { $l_own = $val}
                        elsif ($key eq "dir-group") { $d_grp = $val}
                        elsif ($key eq "file-group") { $f_grp = $val}
                        elsif ($key eq "link-group") { $l_grp = $val}
                    }
                    return 0;
                 }
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as the final part of the wanted_c function. Everything works fine now, except since I am using diagnostics I get:

Variable "%info" will not stay shared at permmy6.pl line 372 (#1)
(W closure) An inner (nested) named subroutine is referencing a
lexical variable defined in an outer subroutine.
    
When the inner subroutine is called, it will probably see the value of
the outer subroutine's variable as it was before and during the *first
+*
call to the outer subroutine; in this case, after the first call to th
+e
outer subroutine is complete, the inner and outer subroutines will no
longer share a common value for the variable.  In other words, the
variable will no longer be shared.
    
Furthermore, if the outer subroutine is anonymous and references a
lexical variable outside itself, then the outer and inner subroutines
will never share the given variable.
    
This problem can usually be solved by making the inner subroutine
anonymous, using the sub {} syntax.  When inner anonymous subs that
reference variables in outer subroutines are called or referenced, the
+y
are automatically rebound to the current values of such variables.
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Any thoughts?

Bother, I was right the first time, and looking at the wrong bit of code the second time.

The wanted_c routine is called once for each file, and in the original code it goes through a series of elsifs until it finds a match, so it will only act on the first match - the first combination of filetype and updatetype that is appropriate.

So to fix it you need to avoid that; the minimal fix would probably be something like this:

  sub wanted_c {
    if (-d $_) {
      if ($d_mode) {
        change_mode($_, $d_mode);
      }
      if ($d_owner) {
        change_owner($_, $d_owner);
      }
      if ($d_group) {
        change_group($_, $d_group);
      }
    } elsif (-l $_) {
      ...
    } elsif (-f $_) {
      ...
    }
  }
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The point is that the different filetypes are exclusive: you're only interested in the first type that matches, so an if/elsif chain is appropriate. The updatetypes are not exclusive: you want to act on each updatetype that is specified, so you much check each of them without jumping out on the first one. An if/elsif chain is not appropriate for that case.

Hugo